We proved the following version of the Hahn-Banach extension theorem in a course I'm taking:
Theorem (Hahn-Banach): Let $X$ be a real vector space and $q : X \to \mathbb{R}$ be sublinear. Let $U \subseteq X$ be a linear subspace
and $l : U \to \mathbb{R}$ be a linear functional that satisfies $l(u)
\leq q(u)$ for all $u \in U$.
Then there exists a linear extension
$L : X \to \mathbb{R}$ of $l$ that satisfies $L(x) \leq q(x)$ for all $x \in X$.
Is it true that this Hahn-Banach extension $L$ is also continuous, hence a $\mathbb{R}$-linear functional?
The way we use the theorem later in this course seems to rely on getting a functional out of the extension. Additionally, the Wikipedia entry states that Hahn-Banach yields a linear functional.
Some of my thoughts/work so far:
I see that, if $q$ is continuous, then $L$ is also continuous:
Let $(x_n) \subseteq X$ with $x_n \to 0$. Then
$$
L(x_n) \leq q(x_n) \to 0
$$
$$
-L(x_n) = L(-x_n) \leq q(-x_n) \to 0
$$
So $|L(x_n)| \leq \max\{q(x_n), q(-x_n)\} \to 0$. Hence, $L$ is continuous at $0$, and therefore continuous.
But a sublinear function isn't necessarily continuous… so this doesn't help, right?
From what I showed above, we see that
$$
-q(-x) \leq L(x) \leq q(x) \qquad \forall x \in X.
$$
Not sure if this inequality might help.
Best Answer
Here is a counterexample: Let $X$ be an infinite-dimensional normed vector space. Then, there is a discontinuous linear functional $q$. Evidently, $q$ is sublinear. Now, take $U = \{0\}$ and $l(0) = 0$. However, the only linear function $L : X \to \mathbb R$ satisfying $q$ is $L = q$ which is discontinuous.